Answer:
The final charges of each sphere are: q_A = 3/8 Q
, q_B = 3/8 Q
, q_C = 3/4 Q
Explanation:
This problem asks for the final charge of each sphere, for this we must use that the charge is distributed evenly over a metal surface.
Let's start Sphere A makes contact with sphere B, whereby each one ends with half of the initial charge, at this point
q_A = Q / 2
q_B = Q / 2
Now sphere A touches sphere C, ending with half the charge
q_A = ½ (Q / 2) = ¼ Q
q_B = ¼ Q
Now the sphere A that has Q / 4 of the initial charge is put in contact with the sphere B that has Q / 2 of the initial charge, the total charge is the sum of the charge
q = Q / 4 + Q / 2 = ¾ Q
This is the charge distributed between the two spheres, sphere A is 3/8 Q and sphere B is 3/8 Q
q_A = 3/8 Q
q_B = 3/8 Q
The final charges of each sphere are:
q_A = 3/8 Q
q_B = 3/8 Q
q_C = 3/4 Q
The equation to find force is f=ma. So, if you plug in the information that you have you'll get F=5x3 and that'll equal F=15N
Answer:
Explanation:
"The thermal energy moving from her coffee to the tongue" represent the heat.
Here coffee is at high temperature while tongue is at low temperature, when Ixchelt tongue make contact with coffee then thermal energy of coffee is absorbed by tongue and tongue gets burned.
As heat always from high Potential to low that is why heat is absorbed by tongue.
Answer:
The number of turns in secondary coil is 4000
Explanation:
Given:
Current in primary coil
A
Current in secondary coil
A
Number of turns in primary coil 
In case of transformer the relation between current and number of turns is given by,

For finding number of turns in secondary coil,



Therefore, the number of turns in secondary coil is 4000
Answer:
Star A is closer than Star B
Explanation:
As we know that in parallax method of distance measurement the angle subtended by the star when it covers a distance of one Parsec arc length, it is known as parallax angle
Here we can say

so we have

so here we have
angle subtended by Star A = 1 arc sec
angle subtended by star B = 0.75 arc sec
now we have
distance for star A is given as

distance of star B is given as

So star A is closer than star B